第三次作业

 参考书《数据压缩导论(第4版)》  Page 100   5, 6

  5、给定如表4-9所示的概率模型,求出序列a1a1a3a2a3a1 的实值标签。

求序列a1a1a3a2a3a的实值标签就是求序列113231的实值标签

        Fx(k)=0, k≤0, Fx(1)=0.2, Fx(2)=0.5, Fx(3)=1, k>3.

       令 l(0)=0, u(0)=1

     1        

               l(1)=  l(0)+(u(0)- l(0))*Fx(0)=0+(1-0)*0=0

                u(1)= l(0)+(u(0)- l(0))*Fx(1)=0+(1-0)*0.2=0.2

    11          

               l(2)=  l(2)+(u(1)- l(1))*Fx(0)=0+(1-0)*0=0

                u(2)= l(1)+(u(1)- l(1))*Fx(1)=0+(0.2-0)*0.2=0.04

   113   

                l(3)=  l(2)+(u(2)- l(2))*Fx(2)=0+(0.04-0)*0.5=0.02

                u(3)= l(2)+(u(2)- l(2))*Fx(3)=0+(0.04-0)*1=0.04

   1132 

                l(4)=l(3)+(u(3)-l(3))*Fx(1)=0.02+(0.04-0.02)*0.2=0.024 

                u(4)=l(3)+(u(3)-l(3))*Fx(2)=0.02+(0.04-0.02)*0.5 =0.03

    11323     

                  l(5)=l(4)+(u(4)-l(4))*Fx(2)=0.024+(0.03-0.024)*0.5=0.027

                  u(5)=(4)+(u(4)-(4))*Fx(3)=0.024+(0.03-0.024)*1=0.03

    113231

            l(6)=l(5)+(u(5)-l(5))*Fx(0)=0.027+(0.03-0.027)*0=0.027

            u(6)=l(5)+(u(5)-(5))*Fx(1)=0.027+(0.03-0.027)*0.2=0.0276

             Tx(113231)= ( u(6) + l(6)   )/2

                               =(0.0276+0.027)/2

                               =0.0273

 6、对于表4-9所示的概率模型,对于一个标签为0.63215699的长度为10的序列进行解码。

     由表4-9可知 Fx(k)=0, k≤0, Fx(1)=0.2, Fx(2)=0.5, Fx(3)=1, k>3.

    令l(0)=0, u(0)=1。设玄素对应的序列为x1x2x3x4x5x6x7x8x9x10  ,则利用更新公式

          x1

                l(1)=  l(0)+(u(0)- l(0))*Fx(x1-1)=Fx(x1-1)

                u(1)= l(0)+(u(0)- l(0))*Fx(x1)=Fx(x1)

               x1=1时,区间为[0,0.2]

               x1=2时,区间为[0.2,0.5]

               x1=3时,区间为[0.5,1]

              因为0.63215699在区间[0.5,1]内,所以,第一个元素的序列为3,元素则为a3

        x2

                  u(2)= l(1)+(u(1)- l(1))*Fx(x2)=0.5+(1-0.5)*Fx(x2)=0.5+0.5Fx(x2)

                 l(2)=l(1)+(u(1)-l(1))*Fx(x2-1)=0.5+(1-0.5)*Fx(x2-1)=0.5+0.5Fx(x2-1)

              x2=1时,区间为[0.5,0.6]

             x2=2时,区间为[0.6,0.75]

             x2=3时,区间为[0.75,1]

            因为 0.63215699在区间[0.6,0.75]内,所以,第二个元素的序列为2,元素则为a2

       x3

                  u(3)= l(2)+(u(2)- l(2))*Fx(x3)=0.6+(0.75-0.6)*Fx(x2)=0.6+0.15Fx(x3)

                  l(3)=  l(2)+(u(2)- l(2))*Fx(x3-1)=0.6+(0.75-0.6)*Fx(x2-1)=0.6+0.15Fx(x3-1)

      

                x3=1时,区间为[0.6,0.63]

                x3=2时,区间为[0.63,0.675]

               x3=3时,区间为[0.675,0.75]

               因为0.63215699在区间[0.63,0.675]内,所以,第三个元素的序列为2,元素则为a2

       x4

                  u(4)=l(3)+(u(3)-l(3))*Fx(x4)=0.63+(0.675-0.63)*Fx(x4)=0.63+0.045*Fx(x4)

                    l(4)=l(3)+(u(3)-l(3))*Fx(x4-1)=0.63+(0.675-0.63)*Fx(x4-1)=0.63+0.045*Fx(x4-1)

     

         x4=1时,区间为[0.63,0.639]

        x4=2时,区间为[0.639,0.6525]

        x4=3时,区间为[0.6525,0.675]

                 因为0.63215699在区间[0.63,0.639]内,所以,第四个元素的序列为1,元素则为a1

    x5

                     u(5)=l(4)+(u(4)-l(4))*Fx(x5)=0.63+(0.639-0.63)*Fx(x5)=0.63+0.009*Fx(x5)

                    l(5)=l(4)+(u(4)-l(4))*Fx(x5-1)=0.63+(0.639-0.63)*Fx(x5-1)=0.63+0.009*Fx(x5-1)

               x5=1时,区间为[0.63,0.6318]

              x5=2时,区间为[0.6318,0.6345]

              x5=3时,区间为[0.6345,0.639]

              因为0.63215699在区间[0.6318,0.6345]内,所以,第五个元素的序列为2,元素则为a2

      x6

                      u(6)=l(5)+(u(5)-l(5))*Fx(x6)=0.6318+(0.6345-0.6318)*Fx(x6)=0.6318+0.0027*Fx(x6)

                     l(6)=l(5)+(u(5)-l(5))*Fx(x6-1)=0.6318+(0.6345-0.6318)*Fx(x6-1)=0.6318+0.0027*Fx(x6-1)

        

              x6=1时,区间为[0.6318,0.63234]

             x6=2时,区间为[0.63234,0.63315]

             x6=3时,区间为[0.63315,0.6345]

               因为0.63215699在区间[0.6318,0.63234]内,所以,第六个元素的序列为1,元素则为a1

      x7

                      u(7)=l(6)+(u(6)-l(6))*Fx(x7)=0.6318+(0.63234-0.6318)*Fx(x7)=0.6318+0.00054*Fx(x7)

                      l(7)=l(6)+(u(6)-l(6))*Fx(x7-1)=0.6318+(0.63234-0.6318)*Fx(x7-1)=0.6318+0.00054*Fx(x7-1)

                 x7=1时,区间为[0.6318,0.631908]

                 x7=2时,区间为[0.631908,0.63207]

                 x7=3时,区间为[0.63207,0.63234]

             因为 0.63215699在区间[0.63207,0.63234],所以,第七个元素的序列为3,元素则为a3

     x8

                      u(8)=l(7)+(u(7)-l(7))*Fx(x8)=0.63207+(0.63234-0.63207)*Fx(x8)=0.63207+0.00027*Fx(x8)

                      l(8)=l(7)+(u(7)-l(7))*Fx(x8-1)=0.63207+(0.63234-0.63207)*Fx(x8-1)=0.63207+0.00027*Fx(x8-1)

                 x8=1时,区间为[0.63207,0.632124]

                 x8=2时,区间为[0.632124,0.632205]

                 x8=3时,区间为[0.632205,0.63234]

               因为0.63215699在区间[0.632124,0.632205]内,所以,第八个元素的序列为2,元素则为a2

     x9

                      u(9)=l(8)+(u(8)-l(8))*Fx(x9)=0.632124+(0.632205-0.632124)*Fx(x9)=0.632124+(8.1e-5)*Fx(x9)

                      l(9)=l(8)+(u(8)-l(8))*Fx(x9-1)=0.632124+(0.632205-0.632124)*Fx(x9-1)=0.632124+(8.1e-5)*Fx(x9-1)

              x9=1时,区间为[0.632124,0.6321402]

              x9=2时,区间为[0.0.6321402,0.6321645]

              x9=3时,区间为[0.6321645,0.63234]

               因为0.63215699在区间[0.6321402,0.6321645]内,所以,第九个元素的序列为2,元素则为a2

     x10

                     u(10)=l(9)+(u(9)-l(9))*Fx(x10)=0.6321402+(0.6321645-0.6321402)*Fx(x10)=0.6321402+(2.43e-5)*Fx(x10)

                      l(10)=l(9)+(u(9)-l(9))*Fx(x10-1)=0.6321402+(0.6321645-0.6321402)*Fx(x10-1)=0.6321402+(2.43e-5)*Fx(x10-1)

                x10=1时,区间为[0.6321402,0.63212886]

                x10=2时,区间为[0.63212886,0.63215325]

                x10=3时,区间为[0.63215325,0.6321645]

              因为0.63215699在区间[0.63215235,0.6321645]内,所以,第九个元素的序列为3,元素则为a3

               所以根据表4-9所给的概率模型得出标签为0.63215699的长度为10的序列的解码结果为:a3 a2 aa1 a2  a1 a3 a2 a2 a3

         

 

posted @ 2015-09-23 15:28  欧金桃  阅读(141)  评论(0编辑  收藏  举报